(4)F(A, B, C, D)=ABC+ABD+(A+B)CD+COD.D 解:F=ABC十ABD+(A+BCD+C由DD =AB(C+D)+AB Cd+Ced D =ABCd+AB CD+CED D =CD+(CD+CD)D CD+C D解: F=ABC +ABD +( A + B) C D + C ⊕ D D =AB ( C+D ) + A B C D + C⊕D D = AB C D + A B C D + C⊕ D D = C D + ( C D + C D ) D = C D +C D = 1